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我做全国1996年联赛试题时,有一个题的答案说：若a~2+b~2=c~2(a,b,c∈R~+),那么c=m~2+n~2(m,n∈R~+)。我当时一开始不明白,就举了几个例子,发现5=1~2+2~2,41=4~2+5~2,10=1~2+3~2,13=2~2+3~2。我仔细观察又发现：(1+2)~2+(2×1×2)~2=5~2,(4+5)~2+(2×4×5)~2=41~2,(2+3)~2+(2×2×3)~2=13~2。然而10却不符合此规律。这时,我发现题中的c是素数,因此联想到大概仅有质数才可能有此规律(因为5,41,13都为质数,而10却是合数),那么(2~2+
When I did the 1996 national league exam, there was a question that said: If a~2+b~2=c~2(a,b,c∈R~+), then c=m~2+n~2 (m,n∈R~+). I didn’t understand at the very beginning. I gave a few examples and found that 5=1~2+2~2,41=4~2+5~2,10=1~2+3~2,13=2~2. +3~2. I observed carefully and found that: (1+2)~2+(2×1×2)~2=5~2,(4+5)~2+(2×4×5)~2=41~2, (2+3)~2+(2×2×3)~2=13~2. However, 10 does not meet this rule. At this time, I discovered that c in the question is a prime number, so it is probably only possible to have this rule (because 5,41,13 are prime numbers, and 10 is a composite number), then (2~2+)